Chapter 5, Problem 5.19P

To determine

Show that when the mass is displaced to a position $\left(x,y\right)$ the potential energy is in the form of $U=\frac{1}{2}k\text{'}{r}^2$ for an anisotropic oscillator. And also determine the constant $k\text{'}$ in terms of $k$ and the corresponding force.

Answer to Problem 5.19P

It is proved that when the mass is displaced to a position $\left(x,y\right)$ the potential energy is in the form of $U=\frac{1}{2}k\text{'}{r}^2$ for an anisotropic oscillator. And the constant $k\text{'}$ in terms of $k$ is $k\text{'}=2k\left(2-\frac{l_0}{a}\right)$ and the corresponding force is $F=-2k\left(2-\frac{l_0}{a}\right)\left(x\widehat{x}+y\widehat{y}\right)$ .

Explanation of Solution

The four spring-mass system is

The mass is displaced a small distance to the position $\left(x,y\right)$,

Write the expression for the potential energy of the spring

    $U=\frac{1}{2}k{x}^2$        (I)

Here, $U$ is the potential energy, $k$ is the spring constant and $x$ is the stretched or compressed length of the spring.

The length of spring 1 is

$d_1=\sqrt{{\left(a-y\right)}^2+x^2}$        (II)

The length of spring 2 is

$d_2=\sqrt{{\left(a-x\right)}^2+y^2}$        (III)

The length of spring 3 is

$d_3=\sqrt{{\left(a+y\right)}^2+x^2}$        (IV)

The length of spring 4 is

$d_4=\sqrt{{\left(a+x\right)}^2+y^2}$        (V)

Given, the mass is displaced to a small distance $\left(x,y\right)$, $x\ll a$ and $y\ll a$. Thus,

$\frac{x}{a}\ll 1$  and $\frac{y}{a}\ll 1$

Write the binomial expansion when $t$ is small

    ${\left(1+t\right)}^{\frac{1}{2}}=1+\frac{1}{2}t-\frac{1}{8}{t}^2+\mathrm{...}$

Rewriting equation (II)

$\begin{array}{c}{d}_1={\left[{\left(a-y\right)}^2+x^2\right]}^{\frac{1}{2}}\\ ={\left[a^2{\left(1-\frac{y}{a}\right)}^2+x^2\right]}^{\frac{1}{2}}\\ =a{\left[{\left(1-\frac{y}{a}\right)}^2+\frac{x^2}{a^2}\right]}^{\frac{1}{2}}\\ =a{\left[\left(1-\frac{2y}{a}+\frac{y^2}{a^2}\right)+\frac{x^2}{a^2}\right]}^{\frac{1}{2}}\end{array}$

Using binomial expansion,

$\begin{array}{l}{d}_1=a{\left[1+\left(-\frac{2y}{a}+\frac{y^2}{a^2}+\frac{x^2}{a^2}\right)\right]}^{\frac{1}{2}}\\ =a\left[1+\frac{1}{2}\left(-\frac{2y}{a}+\frac{y^2}{a^2}+\frac{x^2}{a^2}\right)-\frac{1}{8}{\left(-\frac{2y}{a}+\frac{y^2}{a^2}+\frac{x^2}{a^2}\right)}^2+\mathrm{...}\right]\end{array}$

Given that $x$ and $y$ are small. Thus, neglecting the higher order powers for $x$ and $y$ in the above equation.

$\begin{array}{c}{d}_1\approx a\left[1+\frac{1}{2}\left(-\frac{2y}{a}+\frac{y^2}{a^2}+\frac{x^2}{a^2}\right)-\frac{1}{8}\left(\frac{4y^2}{a^2}\right)\right]\\ \approx a\left[1-\frac{y}{a}+\frac{1}{2}\left(\frac{y^2}{a^2}+\frac{x^2}{a^2}\right)-\frac{1}{8}\left(\frac{4y^2}{a^2}\right)\right]\\ \approx a\left[1-\frac{y}{a}+\frac{y^2}{2a^2}+\frac{1}{2}\frac{x^2}{a^2}-\frac{y^2}{2a^2}\right]\\ \approx a\left[1-\frac{y}{a}+\frac{1}{2}\frac{x^2}{a^2}\right]\end{array}$

$d_1\approx a-y+\frac{1}{2}\frac{x^2}{a}$

The original length of the spring is $l_0$. Thus, the change in the length of the spring is

$\begin{array}{c}{x}_1=d_1-l_0\\ =a-y+\frac{1}{2}\frac{x^2}{a}-l_0\\ =\left(a-l_0\right)-y+\frac{1}{2}\frac{x^2}{a}\end{array}$

The potential energy of spring 1 is

$\begin{array}{c}{U}_1=\frac{1}{2}k{x}_1^2\\ =\frac{1}{2}k{\left[\left(a-l_0\right)-y+\frac{1}{2}\frac{x^2}{a}\right]}^2\\ =\frac{1}{2}k\left[{\left(a-l_0\right)}^2-2y\left(a-l_0\right)+y^2+2\left(a-l_0\right)\left(\frac{1}{2}\frac{x^2}{a}\right)+{\left(\frac{1}{2}\frac{x^2}{a}\right)}^2-2y\frac{1}{2}\frac{x^2}{a}\right]\\ =\frac{1}{2}k\left[{\left(a-l_0\right)}^2-2y\left(a-l_0\right)+y^2+\left(a-l_0\right)\left(\frac{x^2}{a}\right)\right]\end{array}$

Given that $x$ and $y$ are small. Thus, neglecting the higher order powers for $x$ and $y$ in the above equation.

$U_1\approx \frac{1}{2}k\left[y^2-2y\left(a-l_0\right)+x^2\left(1-\frac{l_0}{a}\right)\right]+\frac{1}{2}k{\left(a-l_0\right)}^2+\text{constant}$

$U_1\approx \frac{1}{2}k\left[y^2-2y\left(a-l_0\right)+x^2\left(1-\frac{l_0}{a}\right)\right]+\text{constant}$        (VI)

For $U_2$ , change $y\to x$ and $x\to y$ in equation (VI)

$U_2\approx \frac{1}{2}k\left[x^2-2x\left(a-l_0\right)+y^2\left(1-\frac{l_0}{a}\right)\right]+\text{constant}$        (VII)

For $U_3$ , change $y\to -y$ and $x\to x$ in equation (VI)

$U_3\approx \frac{1}{2}k\left[y^2+2y\left(a-l_0\right)+x^2\left(1-\frac{l_0}{a}\right)\right]+\text{constant}$        (VIII)

For $U_4$ , change $y\to -x$ and $x\to y$ in equation (VI)

$U_4\approx \frac{1}{2}k\left[x^2+2x\left(a-l_0\right)+y^2\left(1-\frac{l_0}{a}\right)\right]+\text{constant}$        (IX)

The total potential energy of the system is

  $U=U_1+U_2+U_3+U_4$

Substitute equation (VI), (VII), (VIII) and (IX) in the above equation to solve for $U$

$\begin{array}{c}U=\frac{1}{2}k\left[y^2-2y\left(a-l_0\right)+x^2\left(1-\frac{l_0}{a}\right)\right]+\text{constant}+\frac{1}{2}k\left[x^2-2x\left(a-l_0\right)+y^2\left(1-\frac{l_0}{a}\right)\right]+\text{constant}\\ +\frac{1}{2}k\left[y^2+2y\left(a-l_0\right)+x^2\left(1-\frac{l_0}{a}\right)\right]+\text{constant}+\frac{1}{2}k\left[x^2+2x\left(a-l_0\right)+y^2\left(1-\frac{l_0}{a}\right)\right]+\text{constant}\\ =\frac{1}{2}k\left[2y^2+2x^2+x^2\left(1-\frac{l_0}{a}\right)+2y^2\left(1-\frac{l_0}{a}\right)\right]+\text{constant}\\ =k\left[\left(x^2+y^2\right)+\left(x^2+y^2\right)\left(1-\frac{l_0}{a}\right)\right]+\text{constant}\\ =k\left(x^2+y^2\right)\left[1+\left(1-\frac{l_0}{a}\right)\right]+\text{constant}\end{array}$ $U=k\left(x^2+y^2\right)\left[2-\frac{l_0}{a}\right]+\text{constant}$        (X)

Since, $r^2=x^2+y^2$ and consider $k\text{'}=2k\left(2-\frac{l_0}{a}\right)$

Equation (X) becomes,

$U=\frac{1}{2}k\text{'}{r}^2+\text{constant}$

Hence it is proved that when the mass is displaced to a position $\left(x,y\right)$ the potential energy is in the form of $U=\frac{1}{2}k\text{'}{r}^2$ for an anisotropic oscillator.

Write the expression for force

    $F=-\nabla U$

$F=-\left(\frac{\partial U}{\partial x}\widehat{x}+\frac{\partial U}{\partial y}\widehat{y}\right)$

$\frac{\partial U}{\partial x}=2xk\left(2-\frac{l_0}{a}\right)$

And,

$\frac{\partial U}{\partial y}=2yk\left(2-\frac{l_0}{a}\right)$

Therefore,

$\begin{array}{c}F=-\left(2xk\left(2-\frac{l_0}{a}\right)\widehat{x}+2yk\left(2-\frac{l_0}{a}\right)\widehat{y}\right)\\ =-2k\left(2-\frac{l_0}{a}\right)\left(x\widehat{x}+y\widehat{y}\right)\end{array}$

Conclusion:

Hence it is proved that when the mass is displaced to a position $\left(x,y\right)$ the potential energy is in the form of $U=\frac{1}{2}k\text{'}{r}^2$ for an anisotropic oscillator. And the constant $k\text{'}$ in terms of $k$ is $k\text{'}=2k\left(2-\frac{l_0}{a}\right)$ and the corresponding force is $F=-2k\left(2-\frac{l_0}{a}\right)\left(x\widehat{x}+y\widehat{y}\right)$ .